2 Definition 7 We define a positive rational number X to be an equivalence class of all pairs y1 equivalent to a fixed pair x1 We denote the set of positive rational numbers Q + Definition 8 Two rational numbers X,Y are equal, denoted X =Y, if there are x1 X Y such that Remark 9 We can easily define X < Y, X Y, X > Y, X Y Theorem 10 Let X, Y be rational numbers Then there is an natural number z such that z X > Y Proof Take x1 X, y1 Y We need to prove that there is z N such that z > y 1 More generally, we prove that for any x, y N, there is z such that zx > y Fix x, Let M 8 {y NP z N, z x > y} 1 M Two cases x=1 Then we have 1 x=x+x>x=1 x 1 Then there is u N such that x =u = 1+u>1 Therefore 1 M y M then y M Since y M there is z N such that z x > y Thus there is u N such that z x= y +u Now consider z x=z x+x= y +u+x It is easy to see that u+ thus there is v N such that u+x=v This gives z x = y + u+x= y + v = y + v > y (4) Thus ends the proof Theorem 11 Define S 8 { X Q + P X x for some x N} Denote X 8 X + 1 Then S satisfies Axioms 1 5 of N and is therefore N 1 by 1 For any X S, define Notation In the following we will stop using capital letters to denote rational numbers Dedekind cuts Definition 12 A subset ξ Q + is called a cut i it contains a rational number, but does not contain all rational numbers; ii every rational number in the set is smaller than every rational number not in the set; iii it does not contain a greatest rational number Exercise 5 Let ξ Q be a cut Prove that [x ξ] [ y < x, y ξ]; [x ξ] [ y > x, y ξ] (5) Exercise 6 Let x Q Then ξ 8 {y QP y < x} is a cut 2

4 Theorem 18 If ξ > η, then there is a unique ζ such that η + ζ = ξ Proof Uniqueness is trivial We prove existence Since ξ > η, the set is not empty Define We first prove that it is a cut i First we prove ζ is not empty A 8 {x Q + P x ξ, x η} (12) ζ 8 {x yp x, y A, x > y} (13) Sincer ξ > η, we have η ξ and thus there is y ξ, y η Now as ξ is a cut, there is x ξ such that x> y Since y is a cut, x> y η x η Thus we have two numbers x, y A with x> y By definition x y ζ ii Next we prove ζ Q + Since ξ is a cut, there is z Q + such that z ξ Now for any w ζ, there are x, y ξ such that w = x y <x <z Thus z ζ iii Then we prove x ζ, y ζ, x < y Assume the contrary, that is x ζ, y ζ but x > y (x = y is ruled out by y ζ) Since x ζ, there are u, v A such that x=u v Thus we have u = x+v > y + v (14) As u ξ is a cut, y +v ξ One the other hand, v η which is a cut implies y +v η since y +v >v Thus we have y + v A But now Contradiction iv Finally we prove that ζ does not have a greatest element y = (y + v) v ζ (15) Take any y ζ By definition there are u, v A such that y = u v Now since ξ is a cut and u ξ, there is w ξ such that w >u Since η is a cut and u η, w η Thus w A and we have ζ x8 w v >u v = y (16) Therefore y is not a greatest element of ζ Definition 19 Denote that ζ in the above theorem as ξ η Multiplication of cuts Theorem 20 Let ξ, η be cuts Then is also a cut ζ 8 {xyp x ξ, y η} (17) 4

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